Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{~kJ} $$ a. How much heat is evolved for the production of \(1.00 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) ? b. How much heat is evolved when \(4.03 \mathrm{~g}\) hydrogen are reacted with excess oxygen? c. How much heat is evolved when \(186 \mathrm{~g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{~L}\) at \(1.0\) atm and \(25^{\circ} \mathrm{C}\). How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Short Answer

Expert verified
a. For the production of 1.00 mole of \(\mathrm{H}_{2}\mathrm{O}(l)\), 286 kJ of heat is evolved. b. When 4.03 g of hydrogen reacts with excess oxygen, 570.57 kJ of heat is evolved. c. When 186 g of oxygen reacts with excess hydrogen, 3323.25 kJ of heat is evolved. d. When the Hindenburg exploded, assuming all of the hydrogen reacted, \(2.33 \times 10^{9} \: \mathrm{kJ}\) of heat was evolved.

Step by step solution

01

Write down the given values

We are given: - Balanced chemical reaction: \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(l)\) - Enthalpy change: \(\Delta H=-572 \: \mathrm{kJ}\) per 2 moles of \(\mathrm{H}_{2}\mathrm{O}(l)\) - We have to find the heat evolved for 1 mole of \(\mathrm{H}_{2}\mathrm{O}(l)\)
02

Calculate the heat evolved for 1 mole of H₂O

Since the balanced reaction equation shows that for every 2 moles of H₂O produced, 572 kJ of heat is evolved, we can determine how much heat evolves for the formation of 1 mole of H₂O simply by dividing the enthalpy change by 2: Heat evolved for 1 mole of H₂O = \(\frac{-572 \: \mathrm{kJ}}{2}\) = \(-286 \: \mathrm{kJ}\) Answer: \(286 \: \mathrm{kJ}\) of heat is evolved for the production of 1.00 mole of \(\mathrm{H}_{2}\mathrm{O}(l)\). b. Calculating the heat evolved when 4.03 g of hydrogen reacts with excess oxygen
03

Find the moles of hydrogen

We are given: - Mass of hydrogen = 4.03 g - Molar mass of hydrogen = 2.02 g/mol (since 1 mole of H₂ contains 2 moles of H atoms) Number of moles of hydrogen = \(\frac{4.03 \: \mathrm{g}}{2.02 \: \mathrm{g/mol}}\) = 1.995 moles
04

Calculate the heat evolved

Using the balanced chemical equation, we can see that 2 moles of hydrogen react to produce 2 moles of water, which means 1 mole of hydrogen reacts to produce 1 mole of water. Therefore, we can directly use the number of moles of hydrogen to find out how much heat is evolved upon the complete reaction: Heat evolved = \(1.995 \: \mathrm{moles} \times (-286 \: \mathrm{kJ/mole}) = -570.57 \: \mathrm{kJ}\) Answer: 570.57 kJ of heat is evolved when 4.03 g of hydrogen reacts with excess oxygen. c. Calculating the heat evolved when 186 g of oxygen reacts with excess hydrogen
05

Find the moles of oxygen

We are given: - Mass of oxygen = 186 g - Molar mass of oxygen = 32.00 g/mol (since 1 mole of O₂ contains 2 moles of O atoms) Number of moles of oxygen = \(\frac{186 \: \mathrm{g}}{32.00 \: \mathrm{g/mol}}\) = 5.8125 moles
06

Calculate the heat evolved

The balanced chemical equation shows that for every 1 mole of O₂ that reacts, 2 moles of H₂O are produced. Therefore, the number of moles of H₂O formed when 5.8125 moles of O₂ react is: Number of moles of H₂O = 2 × Number of moles of O₂ = 2 × 5.8125 = 11.625 moles Now, we can calculate the heat evolved: Heat evolved = \(11.625 \: \mathrm{moles} \times (-286 \: \mathrm{kJ/mole}) = -3323.25 \: \mathrm{kJ}\) Answer: 3323.25 kJ of heat is evolved when 186 g of oxygen reacts with excess hydrogen. d. Calculating the heat evolved when the Hindenburg exploded
07

Convert volume of H₂ gas to moles

We are given: - Volume of H₂ = \(2.0 \times 10^{8} \: \mathrm{L}\) - Pressure = 1.0 atm - Temperature = \(25^{\circ} \mathrm{C}\) = \(298.15 \: \mathrm{K}\) - Ideal gas constant (R) = 0.08206 \(\mathrm{L \cdot atm / (mol \cdot K)}\) Using the Ideal Gas Law, PV = nRT, we can calculate the number of moles of hydrogen (n): \(n = \frac{P \cdot V}{R \cdot T} = \frac{(1.0 \: \mathrm{atm}) \times (2.0 \times 10^{8} \: \mathrm{L})}{(0.08206 \: \mathrm{L \cdot atm / (mol \cdot K}) \times (298.15 \: \mathrm{K})}\) = \(8.15 \times 10^{6} \: \mathrm{moles}\)
08

Calculate the heat evolved

The balanced chemical equation shows that 1 mole of H₂ reacts to produce 1 mole of H₂O. Thus, we can directly use the number of moles of hydrogen to find out how much heat is evolved upon complete reaction: Heat evolved = \(8.15 \times 10^{6} \: \mathrm{moles} \times (-286 \: \mathrm{kJ/mole}) = -2.33 \times 10^{9} \: \mathrm{kJ}\) Answer: \(2.33 \times 10^{9} \: \mathrm{kJ}\) of heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted.

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Most popular questions from this chapter

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

For the following reactions at constant pressure, predict if \(\Delta H>\Delta E, \Delta H<\Delta E\), or \(\Delta H=\Delta E .\) a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

A \(5.00-\mathrm{g}\) sample of aluminum pellets (specific heat capacity \(=\) \(0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity \(=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) are heated to \(100.0^{\circ} \mathrm{C}\). The mixture of hot iron and aluminum is then dropped into \(97.3 \mathrm{~g}\) water at \(22.0^{\circ} \mathrm{C}\). Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta H\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

Consider the following equations: $$ \begin{array}{rlrl} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & & \Delta H=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \Delta H & =-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \mathrm{A} & & \Delta H= & 64.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2 , and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

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